# An equilateral triangle is inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle

**Solution:**

Let ΔOAB be the equilateral triangle inscribed in the parabola y^{2} = 4ax .

Let AB intersect the x-axis at point C

Let OC = k

From the equation of the given parabola, we have

⇒ y^{2} = 4ak

⇒ y = ± 2√ak

Therefore,

the respective coordinates of points A and B are (k, 2√ak) and (k, - 2√ak)

Hence,

AB = AC + CB

= 2√ak + 2√ak

= 4√ak

Since, ΔOAB is an equilateral triangle,

⇒ OA^{2} = AB^{2}

⇒ OC^{2} + AC^{2} = AB^{2}

Therefore,

⇒ k^{2} + (2√ak)^{2} = (4√ak)^{2}

⇒ k ^{2} + 4ak = 16ak

⇒ k ^{2} = 12ak

⇒ k = 12a

Hence,

AB = 4√ak = 4√a × 12a

= 4√12a²

= 8√3 a

NCERT Solutions Class 11 Maths Chapter 11 Exercise ME Question 8

## An equilateral triangle is inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle

**Summary:**

The side of the equilateral triangle inscribed in the parabola y^{2} = 4ax is 8√3 a

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